Problem 21 - University Physics with Modern Physics

Question

A cylinder contains 0.250 mol of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) gas at a temperature of \(27.0^{\circ} \mathrm{C}\) . The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 \(\mathrm{atm}\) on the gas. The gas is heated until its temperature increases to \(127.0^{\circ} \mathrm{C}\) . Assume that the \(\mathrm{CO}_{2}\) may be treated as an ideal gas. (a) Draw a \(p V\) -diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas?(e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 \(\mathrm{atm} ?\)

Step-by-Step Solution

Answer

(a) Isobaric process line. (b) 208.75 J. (c) On the piston. (d) 519.625 J. (e) 728.375 J. (f) 208.75 J.

Step 1: Understanding the Problem

We need to analyze a thermodynamic process where a gas is heated at constant pressure, which involves calculating work done, heat supplied, and changes in internal energy. This involves the use of ideal gas principles and thermodynamic equations.

Step 2: Draw a p-V Diagram (Part a)

Since the process is at constant pressure, the p-V diagram will show isobaric expansion. Since the pressure remains constant, the line on the p-V diagram is a horizontal line swelling to the right as volume increases with temperature.

Step 3: Calculate the Work Done by the Gas (Part b)

For an isobaric process, the work done by the gas can be calculated using the formula: \[ W = p \Delta V \]Since the number of moles \( (n) \) and temperature \( (T) \) change, we use the ideal gas law twice to find the change in volume \( \Delta V \) as:\[ \Delta V = V_2 - V_1 = \frac{nR(T_2 - T_1)}{p} \]Convert temperatures in Celsius to Kelvin:\[ T_1 = 27.0 + 273.15 = 300.15 \, \text{K} \]\[ T_2 = 127.0 + 273.15 = 400.15 \, \text{K} \]Plug in the values (using \( R = 0.0821 \, \text{L atm/mol K} \) and \( p = 1 \, \text{atm} \)):\[ \Delta V = \frac{0.250 \times 0.0821 \times (400.15 - 300.15)}{1} = 2.0525 \, \text{L} \]Therefore,\[ W = 1 \, \text{atm} \times 2.0525 \, \text{L} = 2.0525 \, \text{L atm} \approx 208.75 \, \text{J} \] (using 1 L atm = 101.325 J).

Step 4: Identify What Work is Done On (Part c)

The work done by the gas is exerted on the piston as it moves upward increasing the volume of the cylinder.

Step 5: Calculate the Change in Internal Energy (Part d)

For an ideal gas, the change in internal energy \( \Delta U \) is given by:\[ \Delta U = nC_v \Delta T \]Where \( C_v = \frac{f}{2}R \), f is degrees of freedom for \( CO_2 \), with f = 5 (3 translational and 2 rotational at room temperature) because it's diatomic:\[ C_v = \frac{5}{2} \times 8.314 = 20.785 \text{ J/mol K} \]\[ \Delta U = 0.250 \times 20.785 \times 100 = 519.625 \text{ J} \]

Step 6: Calculate the Heat Supplied to the Gas (Part e)

Using the first law of thermodynamics:\[ Q = \Delta U + W \]\[ Q = 519.625 + 208.75 = 728.375 \text{ J} \]

Step 7: Calculate Work Done at 0.50 atm Pressure (Part f)

Repeat the work done calculation at 0.50 atm:\[ \Delta V = \frac{0.250 \times 0.0821 \times 100}{0.50} = 4.105 \text{ L} \]\[ W = 0.50 \times 4.105 = 2.0525 \text{ L atm} \approx 208.75 \text{ J} \] at 0.5 atm as the change in volume doubles.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a principle that aids in understanding the relationships among pressure, volume, temperature, and the number of moles of an ideal gas. The equation can be represented as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law assumes gases to have perfectly elastic collisions and no volume or attractive forces between particles.

In this exercise, the law is crucial to determine the change in gas volume when temperature changes while the pressure remains constant. By converting temperatures from Celsius to Kelvin, we apply the Ideal Gas Law twice to find initial and final volumes. It is important because understanding this law helps us estimate how gases behave under varying conditions.

Isobaric Process

An isobaric process occurs when the pressure of a system remains constant while other state properties change. In a PV diagram, this is represented by a horizontal line, indicating no change in pressure.

During such a process, like in the provided problem, heating the gas causes it to expand, increasing its volume while keeping the pressure constant. This uniform pressure allows us to easily compute work done by the gas as it causes the piston to move. Understanding isobaric processes is important for analyzing scenarios where heat is transferred under constant pressure, like in many industrial applications.

Internal Energy

Internal energy in a gas depends on its temperature and the type of molecules it contains. For an ideal gas, internal energy change \( \Delta U \) is directly related to its temperature change. The equation \( \Delta U = nC_v \Delta T \) allows us to calculate this change, where \( C_v \) is the molar specific heat at constant volume.

For carbon dioxide, which is considered as diatomic due to its degrees of freedom, the calculation factors include translational and rotational motion. This relationship shows that when temperature increases, the gas's energy rises too, reflecting a gain that participation in energy-related calculations like heat supplied.

Work Done by Gas

In thermodynamics, calculating the work done by a gas provides insights into the energy transfer during processes like expansion. For an isobaric process, the work done \( W \) by the gas can be determined using the formula \( W = p \Delta V \). This implies that work results from the pressure multiplied by the change in volume.

In the problem scenario, heating the gas increases its temperature causing its volume to enlarge, hence the piston moves, performing work. This concept is key in understanding thermodynamic cycles, where energy conversion and work output are core considerations. Recognizing how pressure and volume changes relate to work done is crucial to mastering material behavior under different pressures.